Answer to Question #197960 in Calculus for Dhruv rawat

Question

Find the angle between the curves y^2=ax and ay^2=x^3 (a>0), at the points of intersection other than origin.

Answer 1

$\text{Given, the two curves} y ^2 = ax \space and \space ay^2 = x ^3.\newline \text{Equating the both equations} y ^2 = ax \space and \space ay^2 = x ^3 \text{to get the integrating point,}\newline ax = \frac{x^3}{a}\newline x ^2 = a ^2\newline x = a\newline \text{Then,}\newline y = (a^2) ^{\frac{1}{2 }}= a.\newline \text{The integrating point is } \space (a,a).\newline \text{Differentiate both curves w.r.t} \space x.\newline \text{First curve}\newline y ^2 = ax\newline 2y \frac{dy}{dx}= a\newline \frac{dy}{dx}= \frac{a}{2y}\newline \text{Then, slope of the first curve at } \space (a,a) \space is \space \frac{1}{2}i.e.,m1 =\frac{1}{2}.\newline \text{Now, second curve,}\newline ay^2 = x ^3\newline 2ay\frac{dy}{dx}=3x^2\newline \frac{dy}{dx}= \frac{3}{2}\newline \text{Therefore, } \space m2=\frac{3}{2}.\newline \text{Then, the angle between the curves given by,}\newline tan\theta= | \frac{m1 – m2}{1 + m1*m2} |\newline = | \frac{\frac{1}{2} – \frac{3}{2} }{1 + \frac{1}{2}×\frac{3}{2}}|\newline = |\frac{-4}{7}| \newline =\frac{4}{7}\newline \theta=tan^{-1}(\frac{4}{7})\newline =29.7°\newline \text{Thus, angle is} \space 29.7°.$